Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $n = \dfrac{r^2 + 7r}{-4r^2 - 48r - 140} \times \dfrac{r^2 + 14r + 45}{-5r^2 + 20r} $
Solution: First factor out any common factors. $n = \dfrac{r(r + 7)}{-4(r^2 + 12r + 35)} \times \dfrac{r^2 + 14r + 45}{-5r(r - 4)} $ Then factor the quadratic expressions. $n = \dfrac {r(r + 7)} {-4(r + 5)(r + 7)} \times \dfrac {(r + 5)(r + 9)} {-5r(r - 4)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac {r(r + 7) \times (r + 5)(r + 9) } { -4(r + 5)(r + 7) \times -5r(r - 4)} $ $n = \dfrac {r(r + 5)(r + 9)(r + 7)} {20r(r + 5)(r + 7)(r - 4)} $ Notice that $(r + 5)$ and $(r + 7)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {r\cancel{(r + 5)}(r + 9)(r + 7)} {20r\cancel{(r + 5)}(r + 7)(r - 4)} $ We are dividing by $r + 5$ , so $r + 5 \neq 0$ Therefore, $r \neq -5$ $n = \dfrac {r\cancel{(r + 5)}(r + 9)\cancel{(r + 7)}} {20r\cancel{(r + 5)}\cancel{(r + 7)}(r - 4)} $ We are dividing by $r + 7$ , so $r + 7 \neq 0$ Therefore, $r \neq -7$ $n = \dfrac {r(r + 9)} {20r(r - 4)} $ $ n = \dfrac{r + 9}{20(r - 4)}; r \neq -5; r \neq -7 $